3, $a,$
$17x^2-3x(5x+1)-2x^2=18$
$⇒15x^2-15x^2-3x=18$
$⇒-3x=18$
$⇒x=-6$
Vậy $x=-6$
$b,$
$16x^2-49=0$
$⇒(4x+7)(4x-7)=0$
$+,4x+7=0$
$⇔4x=-7$
$⇔x=\dfrac{-7}{4}$
$+,4x-7=0$
$⇔4x=7$
$⇔x=\dfrac{7}{4}$
Vậy `x∈{±\frac{7}{4}}`
$c,$
$x^2+y^2-2x+4y+5=0$
$⇒x^2-2x+1+y^2+4y+4=0$
$⇒(x-1)^2+(y+2)^2=0(1)$
Vì $(x-1)^2≥0∀x$
$(y+2)^2≥0∀y$
$⇒(1)⇔+,x-1=0$
$⇒x=1$
$+,y+2=0$
$⇒y=-2$
Vậy $x=1;y=-2$